on2t1inf8b

Answered

2022-07-20

The original problem is:

If a, b, c, d are the position vectors of points A, B, C, D respectively such that

$(\overrightarrow{a}-\overrightarrow{d}).(\overrightarrow{b}-\overrightarrow{c})=(\overrightarrow{b}-\overrightarrow{d}).(\overrightarrow{c}-\overrightarrow{a})=0$

then prove that D is the orthocentre of $\mathrm{\Delta}$ ABC.

How do we go about proving that a point is the orthocentre of a triangle? I've tried expanding the dot product but I don't seem to get anywhere.

If a, b, c, d are the position vectors of points A, B, C, D respectively such that

$(\overrightarrow{a}-\overrightarrow{d}).(\overrightarrow{b}-\overrightarrow{c})=(\overrightarrow{b}-\overrightarrow{d}).(\overrightarrow{c}-\overrightarrow{a})=0$

then prove that D is the orthocentre of $\mathrm{\Delta}$ ABC.

How do we go about proving that a point is the orthocentre of a triangle? I've tried expanding the dot product but I don't seem to get anywhere.

Answer & Explanation

kuglatid4

Expert

2022-07-21Added 12 answers

That formula states that $AD\perp BC$ (so that $D$ is on the altitude from $A$ to $BC$) and that $AC\perp BD$ (so that $D$ is on the altitude from $B$ to $AC$). As $D$ is on two altitudes of the triangle, it is its orthocentre.

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